Question Detail

The length of a rectangle is three times of its width. If the length of the diagonal is \begin{aligned}8\sqrt{10}\end{aligned}, then find the perimeter of the rectangle.

60 cm
62 cm
64 cm
66 cm

Answer: Option C

Explanation:

Let Breadth = x cm,
then, Length = 3x cm
\begin{aligned}
x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \\
=> 10x^2 = 640 \\
=> x = 8 \\
\end{aligned}
So, length = 24 cm and breadth = 8 cm
Perimeter = 2(l+b)
= 2(24+8) = 64 cm

1. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.

10 m * 5 m
15 m * 5 m
20 m * 5 m
25 m * 5 m

Answer: Option C

Explanation:

From the question, 2b+l = 30
=> l = 30-2b
\begin{aligned}
Area = 100m^2 \\
=> l \times b = 100 \\
=> b(30-2b) = 100 \\
b^2 - 15b + 50 = 0 \\
=>(b-10)(b-5)=0 \\
\end{aligned}
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m

2. If the radius of a circle is diminished by 10%, then the area is diminished by:

200%
210%
300%
310%

Answer: Option C

Explanation:

Let the original radius be R cm. New radius = 2R

\begin{aligned}
Area = \pi R^2 \\
\text{New Area =} \pi {2R}^2 \\
= 4\pi R^2 \\
\text{Increase in area =}(4\pi R^2 - \pi R^2) \\
= 3\pi R^2 \\
\text{Increase percent =} \frac{3\pi R^2}{\pi R^2}*100 \\
= 300 \%
\end{aligned}

3. The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

15 cm
20 cm
25 cm
30 cm

Answer: Option D

Explanation:

We know the product of diagonals is 1/2*(product of diagonals)

Let one diagonal be d1 and d2
So as per question

\begin{aligned}
\frac{1}{2}*d1*d2 = 150 \\
\frac{1}{2}*10*d2 = 150 \\
d2 = \frac{150}{5} = 30 \\
\end{aligned}

4. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

\begin{aligned} 152600 m^2\end{aligned}
\begin{aligned} 153500 m^2\end{aligned}
\begin{aligned} 153600 m^2\end{aligned}
\begin{aligned} 153800 m^2\end{aligned}

Answer: Option C

Explanation:

Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:

Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]

As per question length is 3x and width is 2x
We know perimeter of rectangle is 2(L+B)
So, 2(3x+2x) = 1600
=> x = 160
So Length = 160*3 = 480 meter
and Width = 160*2 = 320 meter

Finally, Area = length * breadth
= 480 * 320 = 153600

5. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?

Rs. 3430
Rs. 3440
Rs. 3450
Rs. 3460

Answer: Option B

Explanation:

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440